Problem: Simplify the following expression: $y = \dfrac{-8x^2- 23x+3}{-8x + 1}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-8)}{(3)} &=& -24 \\ {a} + {b} &=& &=& {-23} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-24$ and add them together. Remember, since $-24$ is negative, one of the factors must be negative. The factors that add up to ${-23}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${-24}$ $ \begin{eqnarray} {ab} &=& ({1})({-24}) &=& -24 \\ {a} + {b} &=& {1} + {-24} &=& -23 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-8}x^2 +{1}x) + ({-24}x +{3}) $ Factor out the common factors: $ x(-8x + 1) + 3(-8x + 1)$ Now factor out $(-8x + 1)$ $ (-8x + 1)(x + 3)$ The original expression can therefore be written: $ \dfrac{(-8x + 1)(x + 3)}{-8x + 1}$ We are dividing by $-8x + 1$ , so $-8x + 1 \neq 0$ Therefore, $x \neq \frac{1}{8}$ This leaves us with $x + 3; x \neq \frac{1}{8}$.